Lecture 2: Determination od the Molecular Formula of an Unknown
Friday, September 06, 2013
Elemental analysis: complete combustion type C H O →xO y z O 2 2
Use the number of CO and2H O to 2et the molecular formula
Example: 100mg of pure compound yields 275mg of CO and 225mg o2 H O. What is the 2olecular
formula of the compound?
Take the masses of each CO and2H ) and2divide them by their respective molecular weights:
For CO : 275mg/(44mg.mmol )= 6.25mmol
For H 2: 225mg/(18mg/mmol )= 12.5mmol
Calculate the number of moles of both C and H atoms, and divide by the smallest number of moles
In CO 2here is only 1 carbon atom, therefore: 6.25*1 = 6.25
In H2O there are 2 atoms of H, therefore: 12.5*2 = 25
6.25mmol < 25mmol
For C: 6.25/6.25 = 1
For H: 25/6.25 = 4
Therefore we get the following empirical formula: CH 4
Is this the empirical formula or the molecular formula?
Can you double CH ? (4H )*2 →4 H NOT P2SS8BLE!
In this case, the empirical formula is also the molecular formula!
What about the number of O?
Calculate the mass of C and H, then see if it accounts for the original mass.
For C: 6.25mmol * (12.011mg*mmol ) = 75mg1
For H: 25mmol * (1.0079mg*mmol ) = 25mg
Total mass: 75mg+25mg = 100mg
There are no O in the compound
The empirical formula is indeed CH 4
Example: In a given sample, there is 52.2% C and 13.0% H (<100%). What is the molecular formula of
Assume the remaining percentage is O
100-(52.2+13) = 34.8
Calculate the number of moles of C, H and O in 100mg of compound by dividing the mass by the atomic
For C: 52.2g/(12.0g*mol ) = 4.35mol
For H: 13g/(1g*mol ) - 13mol
For O: 34.8g/(16g*mol ) = 2.175mol
Normalize by dividing by the smallest integer: 2.175<4.35<13
For C: 4.35/2.175 = 2
For H: 13/2.175 = 5.977 (round up to 6)
For O: 2.175/2.175 = 1
Therefore we have the following empirical formula: C H O 2 6 In order to get the molecular formula, we need the molecular weight of the compound (using MS).
Rule of 13
If only the molecular weight is known for a given compound, a possible formula can be derived by
dividing the molecular weight into CH units (mass of 13g). The remainder of the molecular weight is
attributed to H.
Example: a compound has a molecular weight of 66g
66 = (5*13)+1
Therefore, we have 5 CH units and 1 H
The formula for the compound is: C H 5 6
Another possible formula can be found by adding heteroatoms and subtracting CH units n
Mass = 16g O or CH
Mass = 14g N or CH 2
Alternative molecular formulas:
C H + O - CH = C H O
5 6 4 4 2
C 5 +6N - CH = C2H N 4 4
Unsaturation Number (UN)
The unsaturation number of an organic compound gives the number of double bonds and/or rings.
General Organic Compound: